College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.4 - Polynomial and Rational Inequalities - 5.4 Assess Your Understanding - Page 373: 53

Answer

The inequality is valid for values less than 3 and values more than 7 (including them) i.e. $(-\infty,3]\cap [7,\infty)$.

Work Step by Step

First, we are going to move everything to the left side and simplify: $\dfrac{x+1}{x-3}\leq2$ $\dfrac{x+1}{x-3}-2\leq0$ $\dfrac{x+1}{x-3}-\dfrac{2(x-3)}{x-3}\leq0$ $\dfrac{x+1-(2x-6)}{x-3}\leq0$ $\dfrac{-x+7}{x-3}\leq0$ Now, we find critical points by equating the numerator and denominator to zero: $x-3=0$ $-x+7=0$ There are two critical points: $x_1-3=0\rightarrow x_1=3$ $-x_2+7=0\rightarrow x_2=7$ Next, we are going to take three values: one less than 3; one between 3 and 7; and one more than 7 to test in the original equation and check if the inequality is true or not: First test with a value less than 3: $\dfrac{2+1}{2-3}\leq2$ $\dfrac{3}{-1}\leq2$ $-3\leq2 \rightarrow \text{ TRUE}$ Second test with a value between 3 and 7: $\dfrac{4+1}{4-3}\leq2$ $\dfrac{5}{1}\leq2$ $5\leq2 \rightarrow \text{ FALSE}$ Third test with a value more than 7: $\dfrac{13+1}{13-3}\leq2$ $\dfrac{14}{10}\leq2$ $1.4\leq2 \rightarrow \text{ TRUE}$ These tests show that the inequality $\dfrac{x+1}{x-3}\leq2$ is valid for values less than 3 and values more than 7 (including them) i.e. $(-\infty,3]\cap [7,\infty)$
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